Given the distribution with xᵢ = 1, 2, 3, ..., n and fᵢ = 1, 2⁻¹, 2⁻², ..., 2⁻⁽ⁿ⁻¹⁾. What is Σxᵢfᵢ equal to?

S = Σᵢ₌₁ⁿ i·2⁻⁽ⁱ⁻¹⁾ = 1 + 2/2 + 3/4 + ... + n/2ⁿ⁻¹. Using the formula for arithmetic-geometric series: S = Σᵢ₌₁ⁿ i·(1/2)ⁱ⁻¹. The sum Σᵢ₌₁^∞ i·xⁱ⁻¹ = 1/(1-x)² for |x|<1. For finite sum, S = (2ⁿ⁺¹ - n - 2)/2ⁿ⁻¹. This is the standard AGP result.